1963 IMO Problems/Problem 5

Problem

Prove that $\cos{\frac{\pi}{7}}-\cos{\frac{2\pi}{7}}+\cos{\frac{3\pi}{7}}=\frac{1}{2}$ .

Solutions

Solution 1

Let $\cos{\frac{\pi}{7}}-\cos{\frac{2\pi}{7}}+\cos{\frac{3\pi}{7}}=S$ . We have

$S=\cos{\frac{\pi}{7}}-\cos{\frac{2\pi}{7}}+\cos{\frac{3\pi}{7}}=\cos{\frac{\pi}{7}}+\cos{\frac{3\pi}{7}}+\cos{\frac{5\pi}{7}}$

Then, by product-sum formulae, we have

$S * 2* \sin{\frac{\pi}{7}} = \sin{\frac{2\pi}{7}}+\sin{\frac{4\pi}{7}}-\sin{\frac{2\pi}{7}}+\sin{\frac{6\pi}{7}}-\sin{\frac{4\pi}{7}}=\sin{\frac{6\pi}{7}}=\sin{\frac{\pi}{7}}$

Thus $S = 1/2$ . $\blacksquare$

Solution 2

Let $a=\sin{\frac{\pi}{7}}$ and $b=\cos{\frac{\pi}{7}}$ . From the addition formulae, we have

$S=b-[b^2-a^2]+[ b(b^2-a^2)-a(2ab) ]=b-b^2+b^3+a^2(1-3b)$

From the Trigonometric Identity, $a^2=1-b^2$ , so

$S=b-b^2+b^3+(1-b^2)(1-3b)=4b^3-2b^2-2b+1$

We must prove that $S=1/2$ . It suffices to show that $8b^3-4b^2-4b+1=0$ .

Now note that $\cos{\frac{4\pi}{7}}=-\cos{\frac{3\pi}{7}}$ . We can find these in terms of $a$ and $b$ :

$\cos{\frac{4\pi}{7}}=[b^2-a^2]^2-[2ab]^2=b^4-6a^2b^2+b^4=b^4-6(1-b^2)b^2+b^4=8b^4-8b^2+1$

$\cos{\frac{3\pi}{7}}=b[b^2-a^2]-a[2ab]=b^3-3a^2b=b^3-3(1-b^2)b=3b-4b^3$

Therefore $8b^4-8b^2+1=-(3b-4b^3)\Rightarrow 8b^4+4b^3-8b^2-3b+1=0$ . Note that this can be factored:

$8b^4+4b^3-8b^2-3b+1=(8b^3-4b^2-4b+1)(b+1)=0$

Clearly $b\neq -1$ , so $8b^3-4b^2-4b+1=0$ . This proves the result. $\blacksquare$

1963 IMO (Problems) • Resources
Preceded by Problem 4	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 6
All IMO Problems and Solutions

Page

Toolbox

Search

1963 IMO Problems/Problem 5

Contents

Problem

Solutions

Solution 1

Solution 2

See Also